Problem 6 (4 points). Let (an) be a sequence in real numbers, and let A be the s
ID: 3281872 • Letter: P
Question
Problem 6 (4 points). Let (an) be a sequence in real numbers, and let A be the set of all terms in an. Let S be the set of all subsequential limits of A = {an : n E N), S = {s :subsequence ank such that lim ank-s) lim sup anp S, lim inf an inf S. (a) S is the set of all accumulation points of A. Namely, S-acc(A). (a), and denote sup S as lim supan and inf S as lim inf a. In symbols, Now prove the following propositions: (Recall that r E acc(A) if and only if for all deleted neighborhood (b) inf A S lim inf an S limsup an sup A. (c) an converges if and only if lim nf a sup an (d) If an converges, then liman = lim infa" = lim sup anExplanation / Answer
(a) if x is in S then every neighborhood of x has infinitely many points of a subsequence and hence of A. So x is accumulation point of A. Thus S is subset of acc(A).
If x is in acc(A) then every neighborhood of x has point of A other than x so we can form a sequence which converge to x. So we have subsequence of A converging to x so, x is in S.
acc(A) is contained in S.
Thus S= acc(A).
(b) liminf an is limit of infimums of subsets of A containing terms having index greater than n. As n increases sequence of infimums increases. Because infimum of subset is greater than or equal to infimum of a set containing it.
Thus infA is less than or equal to liminf an .
for every nonempty set infimum is less than or equal to supremum. so sequence of supremums dominates sequence of infimums and hence liminf is less than or equal to limsup.
supremum of a subset is always greater than its subsets and by definition of limsup we have limsup an is less than or equal to sup A.
(c) we will do it by contraposition.
If an is not convergent then there are two infinite
subsequence coverging to different limits or may be divergent so an can not same liminf and limsup.
If liminf an and limsup an are different we can find two subsequences converging to different limits. So an is not convergent.
(d) If an converges to lim an = a (say) then every neighborhood of a has all but finitely many elements of A so every subsequence of A converges to a . Since sequences of infimums and supremums are subsequences of A they converge to a.